A surface S is given by $x^2+y^2+\frac{z^2}{4}=1, z>0$ $\overrightarrow{F} = (2x+z^3, 3y+x^4+\cos (z^2), z+ \frac{1}{\sqrt{x^2+y^2}})$ The problem is to find the surface integral. Is it possible to use divergence theorem here? I tried to cut regions into two parts where $\overrightarrow{F}$ is $C^1$ so I can apply the divergence theorem, and where I cannot, but it didn't give me any meaningful result.
How should I deal with this? Maybe I guess that the problem is wrong. My teacher didn't expected the discontinuities, and just believed that $\frac{1}{\sqrt{x^2+y^2}}$ vanishes as we take partial derivatives.
Are there any other appropriate method to deal with this?
Add the surface $S_0 = \{ (x,y,z) \in \mathbb R^3 \mid x^2+y^2 \leq 1 \}$.
Then use Divergence Theorem for the region between $S$ and $S_0$. There you have $$\nabla \cdot \vec F = 2 + 3 + 1 = 6.$$
Then on the surface $S_0$ you have $$\vec F = (2x, 3y+x^4+1, \frac{1}{\sqrt{x^2+y^2}})$$ and the normal $\vec n = -\hat z = (0, 0, -1)$ giving $$\vec F \cdot \vec n = -\frac{1}{\sqrt{x^2+y^2}}$$
When you integrate this you can limit the surface to $\sqrt{x^2+y^2}\geq\epsilon$ and then take limit as $\epsilon\to 0$. Use planar polar coordinates!